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Bezzdna [24]
3 years ago
7

Convert the average temperatures for each collected data point given below from °C to K. Plot the average cell potentials E (y-a

xis) vs T (x-axis). The plot should be approximately linear. Add a trendline to find the best linear fit and write down the y-intercept and slope (b and m from the linear equation) for the trendline below.
Average Temperature in °C - Average Cell Potential (V)

15 - 0.465
18 - 0.467
21 - 0.468
24 - 4.69
27 - 0.471
30 - 0.472
33 - 0.474

Chemistry
1 answer:
koban [17]3 years ago
7 0

Answer:

Explanation:

The equation of above line , y = 0.0005x+ 0.458

This can be compared with y = mx+c

Hence slope, m = 0.0005 and Y-intercept, c = 0.458

Or it can be plotted manually where straight line has to be drawn touching maximum number of data points. After drawing a straight linear line, we need to take any two points from the straight line and slope is calculated

Slope,

m = \frac{y_2-y_1}{x_2-x_1}

and y -intercept is calculated using extraplotting backwards such that it touches the Y-axis. the point where straight line touches Y-axis is Y-intercept (c).

Plot the average cell potentials E (y-axis) vs T (x-axis). image attached

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Answer:

B. 3.0 g/ml

Explanation:

density formula: mass/volume

15/5=3

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3 years ago
If an unknown sample contains 39.04% sulfuric acid by mass, then a 0.9368 g of that sample would require _____ mL of 0.2389 M Na
bonufazy [111]

Answer:

A) 31.22

Explanation:

The reaction of sulfuric acid with NaOH is:

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2H₂O

To solve this problem we need to determine the moles of acid that will react, and, using the chemical equation we can determine the moles of NaOH and the volume that a 0.2389M NaOH solution would require to neutralize it.

<em>Moles H₂SO₄ (Molar mass: 98.08g/mol):</em>

0.9368g * 39.04% = 0.3657g H₂SO₄ * (1mol / 98.08g) =

3.7289x10⁻³moles H₂SO₄

And moles of NaOH that you require to neutralize the acid are:

3.7289x10⁻³moles H₂SO₄ * (2 moles NaOH / 1 mole H₂SO₄) =

7.4578x10⁻³ moles NaOH

Using a 0.2389M NaOH solution:

7.4578x10⁻³ moles NaOH * (1L / 0.2389mol) = 0.03122L = 31.22mL

Right answer is:

<h3>A) 31.22 </h3>

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3 years ago
Benzophenone has a normal freezing point of +48.1 oC, with freezing point depression constant Kfpt = − 9.78 oC/m. A 0.1500 molal
Neporo4naja [7]

Answer:

i = 2.79

Explanation:

The excersise talks about the colligative property, freezing point depression.

Formula to calculate the freezing point of a solution is:

Freezing point of pure solvent - Freezing point of solution = m . Kf . i

Let's replace data given. (i = Van't Hoff factor, numbers of ions dissolved in solution)

48.1°C - 44°C = 0.15 m . 9.78°C/m . i

4.1°C / (0.15 m . 9.78°C/m) = i

i = 2.79

In this case, numbers of ions dissolved can decrease the freezing point of a solution, which is always lower than pure solvent.

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3 years ago
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